Question

Solve the equation for all real solutions. 6p^2-5p-6=0

Answer 1

$\text{Use the quadratic formula:}\\\\\\p=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\text{Plug in and solve:}\\\\p=\frac{-(-5)\pm\sqrt{-5^2-4(6)(-6)}}{2(6)}\\\\p=\frac{5\pm\sqrt{25+144}}{12}\\\\p=\frac{5\pm\sqrt{169}}{12}\\\\p=\frac{5\pm13}{12}\\\\\text{Solve for addition:}\\\\p=\frac{5+13}{12}\\\\p=\frac{18}{12}\\\\\boxed{p=\frac{3}{2}}\\\\\text{Solve for subtraction:}\\\\p=\frac{5-13}{12}\\\\p=\frac{-8}{12}\\\\\boxed{p=-\frac{2}{3}}$

Answer 2

Answer:

p= -2/3 or p=3/2

Step-by-step explanation:

(3p +2)(2p -3)=0

3p=-2 or 2p=3

p=-2/3 or p=3/2